Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following question:

If $(x^1,\ldots,x^n)$ are the standard coordinates on $R^n$ and $(y^1,\ldots,y^n)$ are other coordinates on $R^n$, how can we show that $\sum x^i\frac{\partial}{\partial x^i} =\sum y^i\frac{\partial}{\partial y^i}$ ?

Thank you!

share|improve this question
    
Welcome to MSE =) –  Patrick Da Silva Nov 20 '11 at 20:34
add comment

2 Answers

In the title of your question you mention the Euler vector field, so I suspect that your question was about the following fact:

for any vector bundle $\pi:E\to M$, the Euler vector field $X$ on $E$, i.e. the infinitesimal generator of the homotheties of $E$, has the expression $X=u\frac{\partial}{\partial u}$ w.r.t. standard coordinates $(x,u)$.

If I have understood your question then below there is my answer.


If $\pi:E\to M$ is a vector bundle then its Euler vector field $X$ is defined as the vector field on $E$ which is the infinitesimal generator of the action $\Phi$ of $\mathbb{R}$ on $E$ given by the fiberwise scalar multiplication.
$$\Phi:(t,e)\in\mathbb{R}\times E\to\Phi_t(e):=\exp(t).e\in E.$$ $$X_e:=\left.\frac{d}{dt}\right|_{t=0}\ \Phi_t(e),\ \forall e\in E.$$

Let $x:U\to\mathbb{R}^n$ be coordinates on $M$ and $(x,u):\pi^{-1}(U)\to\mathbb{R}^n\times\mathbb{R}^n$ the associate standard coordinates on $E$. In such coordinates we find the following explicit expressions: $$\Phi(t,(x,u))=(x,\exp(t).u),$$ $$X_{(x,u)}=\left.\frac{d}{dt}\right|_{t=0}\ (x,\exp(t).u)=\left.u\frac{\partial}{\partial u}\right|_{(x,u)}.$$

share|improve this answer
add comment

$y$'s have to be linear functions of $x$'s, otherwise the identity will not hold. If they are, notice that $(\sum x^i \partial_{x^i})f=f$ for any linear function $f$, in particular $(\sum x^i \partial_{x^i})y^k=y^k$, which gives the components of the vector field in the $y$-coordinates.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.